Derivation of Friction Equations
About 9 min read
| There are a number of applications of friction, such as stopping a vehicle in motion and starting up a stationary object. You need to know some basic Physics principles and use some logical thinking to derive the equations for the various problems. The following sections provide basics and is results in a class of interesting problems and solutions.
1. Basic friction equationsThe resistive force of friction (or the force required to overcome friction) between two objects is: Fr = fr x N where fr is the coefficient of friction for the two materials and N is the normal or perpendicular force pushing the two objects together. CoefficientsIf there is no motion, fr is the static coefficient, and if there is motion, fr is the kinetic coefficient. Normal forceThe normal force N can be an external force, such as used on brake pads to stop a vehicle, or it can be caused by the weight of one object on the other. If N is a result of gravity, N = m x g Fr = fr x m x g where m is the mass of the top object and g is the acceleration of gravity (32 ft/sec/sec or 9.8 meters/sec/sec). Newton’s force equationNewton’s force equation is used to determine the force required to accelerate or decelerate and object. When a force is applied to a stationary object, the object will accelerate. If the force is applied to a moving object in the direction of motion, the object will accelerate, but if the force is in the opposite direction, the object will decelerate or slow down. Newton’s equation for the relationship between force and acceleration is: F = m x a where F is the force, m is the mass, and a is the acceleration (or deceleration). AccelerationAcceleration is the change in velocity over a period of time. a = (v2 – v1) / t where v1 is the initial velocity and v2 is the final velocity in acceleration. They are reversed for deceleration. If the velocity starts at v and ends at 0, the deceleration is: a = v / t 2. Stopping a sliding objectSuppose you slide a box along the floor. How long would it travel before it stopped? This would also apply to slamming on the brakes of a moving automobile and measuring how far it would skid. The force decelerating the object equals the force of friction. F = Fr F = m x a = Fr a = Fr / m where a = deceleration of the moving object and m = the mass of the moving object. Since the object is going from v to 0, a = v / t TimeThus, the time it takes to stop the sliding object is: t = m x v / Fr But since momentum P = m x v, the time it takes the object to stop can be written as: t = P / Fr This is a compact and interesting expression. It says that the time it takes a car to skid to a stop is the car’s momentum divided by the friction on the tires. DistanceNow, distance = velocity x time, but since the velocity goes from v to 0, we must take the average velocity. d = (v / 2) x t Thus, the distance for the object to stop d = (m x v^2) / (2 x Fr) where v^2 is v squared. But since energy E = m x v^2 / 2, so d = E / Fr This is also an interesting equation. It says that the distance a car would skid after slamming on its brakes would be the energy of the moving car divided by the friction of the tires. The combination of the equations for time and distance is fascinating. Force of weightIf the normal force N is caused by the weight of the top object, then
so that t = v / (fr x g) d = v^2 / (2 x fr x g) 3. Starting a car by spinning its tiresTorque on the axle of an automobile wheel causes it to turn. If the static friction of the tires on the road is not sufficient to move the car forward and the torque is great enough, the tires will start to spin. We want to know how long it will take for the tires to stop spinning and the car to move forward at a given speed. Given
Find
DerivationThe force of friction would accelerate the car forward until its velocity reached the velocity of the tire surface, at which time the tires would stop spinning and simply roll along, pulling the car forward. If you look at this as relative motion, the problem is the same as the one of stopping a sliding object. The tire surface is sliding and until it stops spinning on the road. The derivation is exactly the same as the previous. Thus the time it takes the tires to stop spinning is: t = v / (fr x g) The distance the car will go until the tires stop spinning is: d = v^2 / 2(fr x g) 4. Moving an object by moving its platformIf you quickly pull a table cloth from under the dishes on the table, the dishes will move slightly, depending on the speed and amount of friction. Likewise, if a box was placed on a conveyor belt, it would slide for a while, until it got up to the speed of the belt. Again, if you look at the relative motion, you can see that this is actually the same situation as the first sliding object problem, except that v is now the velocity of the conveyor belt instead of the object. The equations for t and d are the same as above. 5. Moving a platform by sliding an objectSuppose you had a platform sitting on some surface or even wheels, and you tried to move that platform by sliding another object along its top. You want to find out how long it will take for the platform to reach the speed of the driving object. Given
Find
DerivationThe force on the platform is the difference of the two friction forces: F = Fr – Fp = m x a Fr = fr x N = fr x m x g a = v / t fr x m x g – Fp = m x v / t Thus, t = (m x v) / ((fr x m x g) – Fp) and d = (m x v^2) / 2((fr x m x g) – Fp) 6. Moving a platform by spinning a wheelInstead of moving the platform by sliding an object across its upper surface, suppose you drove the platform with a wheel. You want to know how long the wheel will have to spin until the platform reaches its speed. This is similar to the previous derivation, except that v is the speed of the outer edge of the wheel on the platform. The equations are the same as before. 7. Equations when the normal force is not weightIn many cases, the normal force is the weight of the object. But there are situations–like stopping a wheel from turning–where the force is an applied force. One coefficient of frictionThe equations for t and d for stopping a sliding object, starting a car by spinning its tires, and moving an object by moving its platform are: t = (m x v) / (fr x N) d = (m x v^2) / 2(fr x N) where N is the normal force pressing the objects together. Since momentum of the moving object is P = m x v and its energy is E = m x v^2 /2, these equations can be rewritten as: t = P / (fr x N) d = E / (fr x N) Two coefficients of frictionThe equations for t and d for moving a platform by sliding an object and moving a platform by spinning a wheel are:t = ( v x m) / (N x (fr – fp)) d = (v^2 x m) / (2N x (fr – fp)) where fp is the coefficient of friction between the platform and the second surface and N is the normal force pressing the objects together. Assume that weight is not a factor and that the pressure from the top object or wheel is transferred to the bottom. In conclusionThe above derivations of friction equations should help in solving related problems. Follow a logical approach to solving problems |
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